Question

A particle starts from rest at time t=0 and moves in a straight the w a
acceleration a(ms-2). Find the time at which the (a) speed of the particle will be
4 times its speed at t=2s and (b) distance covered by it will be 8 times that
travelled by the particle at t = 2 s.
​

Answer 1

V=u+at

initially the particle is at rest so u=0

therefore velocity at t=2 is v=a*2

1) now initial velocity is 2a and final velocity is 8a

so for time,

u=0

8a=0+at

therefore t=8

2) apply formula s=ut+(1/2)a*t^2

distance travelled at t=2s

u=0

s=(1/2)a*4

therefore s=2a

so the time taken for the body to travel 8 times the distance travelled at t=2s is

s=ut + (1/2)a*t^2

u=0

so,

16a= (1/2)a*t^2

t^2=32

therefore t=5.66

Answer 2

v=u+at

Initially,the particle is at rest so, u=o

therefore,velocity at t=2,v=a×z

(i)Now,initial velocity is 2a and final velocity is 8a

So,for time

u=0

8a=0+at

therefore t= 8

(ii)apply formula s=ut+(1/2)a×t^2

distance travelled t=2s

4=0

s=(1/2)u×4

therefore s= 2a

So the time taken for the body to travel 8 times the distance travelled t=2s is

su=ut+(1/2)at^2

y=0

So,

16a=(1/2)r×t^2

t^2=32

therefore t=5.66

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